Open Does Not Mean Closed
An open ball is open in the topology defined by its metric. That fact does not make it closed: openness and closedness are separate properties, and a set may be open, closed, both, or neither. The standard real line gives a simple example of an open ball that is not closed. The discrete metric gives the contrasting case: every open ball is both open and closed.
For a metric space \((X,d)\), an open ball with centre \(x\in X\) and radius \(r>0\) is \(B_d(x,r)=\{y\in X:d(x,y)<r\}\). The metric induces a topology in which a set is open if every one of its points lies in some open ball contained in that set. A set is closed if its complement is open. In particular, proving that a ball is not closed means showing that its complement is not open.
Open Balls in the Standard Real Line
Throughout this section, \(\mathbb{R}\) has its standard topology, induced by the metric \(d(x,y)=|x-y|\). For any \(a\in\mathbb{R}\) and \(r>0\), the ball condition is \(|x-a|<r\), which is equivalent to \(a-r<x<a+r\). Thus \(B(a,r)=(a-r,a+r)\).
Proof. Fix \(a\in\mathbb{R}\) and \(r>0\). The ball is the interval \((a-r,a+r)\), which is open in the standard topology. To show it is not closed, we show that its complement is not open. The point \(a+r\) belongs to the complement, since the ball condition is strict: \(|(a+r)-a|=r\), which is not less than \(r\). Let \(V\) be any open neighborhood of \(a+r\). Because \(V\) is open in the standard topology, there is a \(\delta>0\) such that \((a+r-\delta,a+r+\delta)\subseteq V\). Set \(t=a+r-\min(\delta,r)/2\). Then \(0<\min(\delta,r)/2<\delta\), so \(t\in V\). Also \(t<a+r\), while \(t>a\) because \(\min(\delta,r)/2\le r/2<r\). Hence \(a-r<t<a+r\), and therefore \(t\in B(a,r)\). Every open neighborhood of \(a+r\) consequently meets the ball, so no such neighborhood is contained in the complement. The complement is not open, and the ball is not closed. \(\square\)
The endpoint matters because the inequality defining a ball is strict: points at distance exactly \(r\) are excluded. But an excluded endpoint is not separated from the interval by a gap. Every open neighborhood of that endpoint contains points whose distance from the centre is less than \(r\).
Worked Example: A Ball with Integer Endpoints
In \(\mathbb{R}\) with the standard metric, calculate \(B(1,3)\), and check directly that it is open but not closed.
By definition, \(x\in B(1,3)\) exactly when \(|x-1|<3\). This inequality is equivalent to \(-3<x-1<3\). Adding \(1\) throughout gives \(-2<x<4\), so \(B(1,3)=(-2,4)\). The endpoints are not included: at \(x=-2\), \(|-2-1|=3\), and at \(x=4\), \(|4-1|=3\); neither distance is strictly less than \(3\).
The interval \((-2,4)\) is open in the standard topology. Its complement in \(\mathbb{R}\) is \((-\infty,-2]\cup[4,\infty)\). The point \(4\) is in this complement. For any \(\delta>0\), let \(t=4-\min(\delta,3)/2\). Then \(4-\delta<t<4\), so \(t\in(4-\delta,4+\delta)\), while \(1<t<4\), so \(t\in(-2,4)\). Thus every open neighborhood of \(4\) meets the ball. The complement is not open, and the ball is not closed.
Check: Substituting the computed endpoints into the distance formula gives \(|-2-1|=3\) and \(|4-1|=3\), exactly the radius. This confirms that both endpoints are excluded by the strict inequality.
Discrete-Metric Balls Are Clopen
Now let \(X\) be any nonempty set with the discrete metric \(d(x,y)=0\) when \(x=y\), and \(d(x,y)=1\) when \(x\ne y\). Its open balls depend on whether the radius reaches the distance \(1\) between distinct points. The strict inequality in the definition means the case \(r=1\) belongs with radii below \(1\), not above it.
Proof. Fix \(x\in X\) and \(r>0\). The distance from \(x\) to itself is \(0<r\), so \(x\in B_d(x,r)\). If \(y\ne x\), then \(d(x,y)=1\). Such a point belongs to the ball exactly when \(1<r\). Therefore, if \(0<r\le1\), no point other than \(x\) belongs to the ball, and the ball is \(\{x\}\). If \(r>1\), every point of \(X\) belongs to the ball, so the ball is \(X\).
It remains to check openness and closedness. Every subset \(A\subseteq X\) is open in the topology induced by this metric: for each \(z\in A\), the ball \(B_d(z,1/2)=\{z\}\) is contained in \(A\). Applying the same argument to \(X\setminus A\) shows that the complement is open, so \(A\) is also closed. This includes the empty set and \(X\) itself. In particular, every ball described above is clopen. \(\square\)
Worked Example: The Threshold Radius in a Three-Point Space
Let \(X=\{p,q,s\}\) have the discrete metric. Find the balls centred at \(p\) with radii \(1\) and \(3/2\), and check their openness and closedness.
For radius \(1\), the distance calculations are \(d(p,p)=0\), \(d(p,q)=1\), and \(d(p,s)=1\). Membership requires distance strictly less than \(1\). Thus \(p\) is included because \(0<1\), while \(q\) and \(s\) are excluded because \(1<1\) is false. Therefore \(B_d(p,1)=\{p\}\).
For radius \(3/2\), the same distances satisfy \(0<3/2\) and \(1<3/2\). All three points belong, so \(B_d(p,3/2)=\{p,q,s\}=X\). To check the topology directly, every point \(z\in X\) has \(B_d(z,1/2)=\{z\}\). These singleton balls show that \(\{p\}\) is open and that its complement \(\{q,s\}\) is open. The whole set \(X\) is open, and its complement \(\varnothing\) is open. Both balls are therefore closed as well as open.
Check: At radius \(1\), the points at distance exactly \(1\) are excluded; at radius \(3/2\), they are included because \(1<3/2\). This verifies both cases using the ball definition.
A Ball in a Metric Subspace
The same distance formula can produce a different-looking interval when the underlying space is restricted. For a subset of a metric space, the restricted metric measures distances as before, but a ball contains only points of that subset. Closedness must likewise be tested relative to that space: the complement is taken inside the space, not in all of \(\mathbb{R}\).
Worked Example: A Ball in the Unit Interval
Let \(X=[0,1]\) with the metric \(d(x,y)=|x-y|\), and find \(B_d(0,1/2)\). Determine whether it is closed in \(X\).
A point \(x\in X\) belongs to the ball exactly when \(|x-0|<1/2\). Since \(x\in[0,1]\), this is equivalent to \(0\le x<1/2\), so \(B_d(0,1/2)=[0,1/2)\). To verify openness directly, take any \(x\in[0,1/2)\) and set \(s=(1/2-x)/2\), which is positive. If \(y\in X\) and \(d(x,y)<s\), then \(y<x+s=(x+1/2)/2<1/2\), while \(y\ge0\) because \(y\in X\). Hence \(y\in[0,1/2)\), and \(B_d(x,s)\) is contained in the ball.
The complement in \(X\) is \([1/2,1]\). It contains \(1/2\). For any \(\delta>0\), choose \(t=1/2-\min(\delta,1)/2\). Then \(0\le t<1/2\) and \(|t-1/2|<\delta\), so \(t\in X\) lies in the relative ball \(B_d(1/2,\delta)\) but not in the complement. Thus the complement is not open in \(X\), and \([0,1/2)\) is not closed there.
Check: The point \(1/2\) is excluded from the ball because its distance from \(0\) equals the radius. The complement is taken in \(X\), so it is \([1/2,1]\), not a subset of the whole real line.
A Common Mistake and the Main Distinction
A common mistake is to omit the endpoints when taking the complement of an open interval. For example, a wrong calculation is \(\mathbb{R}\setminus(-2,4)=(-\infty,-2)\cup(4,\infty)\). That expression leaves out both \(-2\) and \(4\). The correct complement is \((-\infty,-2]\cup[4,\infty)\). The omitted endpoints are precisely the points that prevent the complement from being open.
There is a related mistake in the opposite direction: concluding that no open ball can be closed because the real-line example is not closed. The discrete-metric calculation disproves that claim. In that metric, a ball is either a singleton or the whole space, and both it and its complement are open. Whether an open ball is closed depends on the metric and the space, not just on the word “open.”
The distinction is useful whenever a set is described by a strict distance inequality. The inequality identifies which points belong to the ball, but closedness asks a separate question about the complement in the chosen topology. In the standard real line, the excluded endpoints can be approached from inside the interval. In the discrete metric, each point has a neighborhood containing no other point, so this endpoint phenomenon does not arise.
Check Your Understanding
For each question, pay attention to the metric, the strict inequality in the ball definition, and the space in which the complement is taken.
- In standard \(\mathbb{R}\), write \(B(-1,2)\) as an interval and explain why it is not closed.
- In a discrete metric space, what is \(B_d(x,1)\)? What changes when the radius is \(5/4\)?
- Why does showing that \(B_d(x,r)\) is open not by itself show that it is closed?
- In \(X=[0,1]\) with the restricted standard metric, compute \(B_d(1,1/3)\) and identify its complement in \(X\).
- Give the correct complement of \((a-r,a+r)\) in \(\mathbb{R}\), for \(r>0\), and identify the points that show it is not open.